Gravitation Test

Result

Gravitation Test - 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

What is the percentage decrease in the weight of a body when it is taken 32 km below the surface of the earth? [Radius of the earth=6400km]

A
1%

B
0.75 %

C
0.5 %

D
0.25 %

Solution
Question 2
1 / -0

The value of $$'g'$$ at a depth of $$80 km$$ will be (Radius of earth $$= 6400 km$$ and value of $$'g'$$ on the surface of earth is $$10m/s^2$$)

A
$$990 \ cm/s^2$$

B
$$980 \ cm/s^2$$

C
$$970 \ cm/s^2$$

D
$$1000 \ cm/s^2$$

Solution

Let $$g'/ge$$ the acceleration due to gravity in mins
Then $$g'=g\left(1\times \cfrac {d}{R}\right)=9.8\left(1-\cfrac {640}{6400}\times 10^3\right)$$
$$=9.8 \times 0.9999m/s^2$$
$$=9.799m/s^2$$
Question 3
1 / -0

The value of gravitational constant depends upon :

A
temperature of the atmosphere

B
masses

C
distance between the masses

D
none of these

Solution

The gravitational constant is a universal constant. It remains as it is in any condition.
It does not depend on any of the factors like the temperature, mass, distance between the masses.

Value of G is $$ 6.67 \times 10^{-11} \, Nm^2 kg^{-2} $$
Question 4
1 / -0

A body has weight (W) on the ground. The work which must be done to lift it to a height equal to the radius of the earth, R, is

A
equal to W $$\times$$ R

B
greater than W $$\times$$ R

C
less than W $$\times$$ R

D
Cannot be determined

Solution

If acceleration due to gravity was constant, constant force W was to be applied.
However, due to the decrease in g we need to apply less force as we go higher.
Hence the work done is lesser.
Question 5
1 / -0

The ratio of the radius of a planet A to that of planet B is $$r$$. The ratio of accelerations due to gravity for the two planets is $$x$$. The ratio of the escape velocities from the two planets is :

A
$$\sqrt{rx}$$

B
$$\sqrt{\dfrac{r}{x}}$$

C
$$\sqrt{r}$$

D
$$\sqrt{\dfrac{x}{r}}$$

Solution

Escape Velocity is given by: $$V_{e}=\sqrt{2gR}$$
Given: $$ \dfrac{R_A}{R_B} = r$$ and $$\dfrac{g_A}{g_B} = x $$
Taking ratio of the escape velocities of the $$2$$ planets:
$$\dfrac{V_A}{V_B} = \sqrt{rx}$$
Question 6
1 / -0

Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by

A
A goes on decreasing and that of B goes on increasing

B
B goes on decreasing and that of A goes on increasing

C
each decreases at the same rate

D
each decreases at different rates

Solution

The variation of 'g' with altitude is: $$g' = g(1- \dfrac{2h}{R})$$
The variation of 'g' with depth is: $$g' = g(1- \dfrac{d}{R})$$
So, from the above equations we can deduce that if a person goes deep into a mine or to a height above the Earth's surface, the gravitational field decreases at different rates.
Question 7
1 / -0

The ratio of acceleration due to gravity at a depth $$h$$ below the surface of earth and at a height $$h$$ above the surface for $$h<<R$$

A
is constant.

B
increases linearly with h.

C
varies parabolically with h.

D
decreases.

Solution

Acceleration due to gravity at a depth $$h$$ below the surface of earth is
$$g_d = g(1 - \dfrac{h}{R})$$
Acceleration due to gravity at a height $$h$$ above the surface of earth is
$$g_h = g(1- \dfrac{2h}{R})$$
where, $$g$$ is the acceleration due to gravity on the surface and $$R$$ is the radius of earth.
So, $$\dfrac{g_d}{g_h} = \dfrac{(R - h)}{ (R - 2h)}$$
Since $$R>> h$$
We can conclude that this ratio is constant.
Question 8
1 / -0

If $$R$$ is radius of the earth and $$W$$ is work done in lifting a body from the ground to an altitude $$R$$, the work which should be done in lifting it further to twice that altitude is:

A
$$\dfrac{W}{2}$$

B
$$W$$

C
$$\dfrac{W}{3}$$

D
$$3W$$

Solution

Work done is equal to change in PE.

$$W = - \Delta U = U_1-U_2 = -\dfrac{GMm}{R}-(- \dfrac{GMm}{2R}) = -\dfrac{GMm}{2R}$$

Work done in lifting body from $$2R$$ to $$4R$$.

$$W' = - \Delta U = U_1-U_2 = -\dfrac{GMm}{2R}-(- \dfrac{GMm}{4R}) = -\dfrac{GMm}{4R} =\dfrac{W}{2}$$
Question 9
1 / -0

A space craft is launched in a circular orbit very close to the earth. What additional velocity should be given to the space craft so that it might escape the earth's gravitational pull:

A
$$20.2 Kms^{-1}$$

B
$$3.25 Kms^{-1}$$

C
$$8 Kms^{-1}$$

D
$$11.2 Kms^{-1}$$

Solution

Since the spacecraft is moving very close to the surface of earth, we assume its orbital radius to be R(the radius of the earth); implying its current orbital velocity is: $$V_{o}\, = \sqrt{gR}$$.
But by definition of escape velocity, we know that to escape the surface of earth, the speed required is $$V_{e}\, = \sqrt{2gR}$$.
Hence, the additional speed to be given to the spacecraft is $$V_{e}\, -\, V_{o}$$.
Using $$g=9.81$$ and $$R=6400\ km$$, we get the required additional speed as $$3.25km\, s^{-1}$$.
Question 10
1 / -0

A particle hanging from a spring stretches it by $$1 cm$$ at earth's surface. Radius of earth is $$6400 km$$. At a place $$800 km$$ above the earths surface, the same particle will stretch the spring by

A
$$0.79 cm$$

B
$$1.2 cm$$

C
$$4 cm$$

D
$$17 cm$$

Solution

as, $$mg=kx$$
where, $$x=1cm, k=$$spring constant
for second case $$mg'=kx'$$
Hence $$ \dfrac{g}{g'}=\dfrac{x}{x'}$$
$$x'=xg'/g$$
$$x'=\dfrac{xgR^2}{(R+h)^2 g}$$
Earth's radius, $$R=6400$$
At height, $$h=800$$
$$x'=\dfrac{1\times (6400 \times 10^3)^2}{(7200 \times 10^3)^2}$$
$$x'=0.79 cm$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Gravitation Test - 24

Result

Gravitation Test - 24

Score

-

Rank

-

SHARING IS CARING

Question 1

1%

0.75 %

0.5 %

0.25 %

Solution

Question 2

$$990 \ cm/s^2$$

$$980 \ cm/s^2$$

$$970 \ cm/s^2$$

$$1000 \ cm/s^2$$

Solution

Question 3

temperature of the atmosphere

masses

distance between the masses

none of these

Solution

Question 4

equal to W $$\times$$ R

greater than W $$\times$$ R

less than W $$\times$$ R

Cannot be determined

Solution

Question 5

$$\sqrt{rx}$$

$$\sqrt{\dfrac{r}{x}}$$

$$\sqrt{r}$$

$$\sqrt{\dfrac{x}{r}}$$

Solution

Question 6

A goes on decreasing and that of B goes on increasing

B goes on decreasing and that of A goes on increasing

each decreases at the same rate

each decreases at different rates

Solution

Question 7

is constant.

increases linearly with h.

varies parabolically with h.

decreases.

Solution

Question 8

$$\dfrac{W}{2}$$

$$W$$

$$\dfrac{W}{3}$$

$$3W$$

Solution

Question 9

$$20.2 Kms^{-1}$$

$$3.25 Kms^{-1}$$

$$8 Kms^{-1}$$

$$11.2 Kms^{-1}$$

Solution

Question 10

$$0.79 cm$$

$$1.2 cm$$

$$4 cm$$

$$17 cm$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.